Question

Suppose an experiment leads to the following system of equations:

\(\begin{aligned}{c}{\bf{4}}.{\bf{5}}{x_{\bf{1}}} + {\bf{3}}.{\bf{1}}{x_{\bf{2}}} = {\bf{19}}.{\bf{249}}\\1.6{x_{\bf{1}}} + 1.1{x_{\bf{2}}} = 6.843\end{aligned}\) (3)

1. Solve system (3), and then solve system (4), below, in which the data on the right have been rounded to two decimal places. In each case, find the exactsolution.

\(\begin{aligned}{c}{\bf{4}}.{\bf{5}}{x_{\bf{1}}} + {\bf{3}}.{\bf{1}}{x_{\bf{2}}} = {\bf{19}}.{\bf{25}}\\1.6{x_{\bf{1}}} + 1.1{x_{\bf{2}}} = 6.8{\bf{4}}\end{aligned}\) (4)

2. The entries in (4) differ from those in (3) by less than .05%. Find the percentage error when using the solution of (4) as an approximation for the solution of (3).

3. Use your matrix program to produce the condition number of the coefficient matrix in (3).

Short answer

1. The exact solution of (3) is \({x_1} = 3.94\) and \({x_2} = 0.49\). The exact solution of (4) is \({x_1} = 2.90\) and \({x_2} = 2.00\).
2. The percentage error in\({x_1} = 2.90\)when using the solution of (4) as an approximation for the solution of (3) is\(26\% \). The percentage error in\({x_2} = 2.00\)when using the solution of (4) as an approximation for the solution of (3) is\(308\% \).
3. The condition number of the coefficient matrix is 3363.

Step-by-step solution

(a) Step 1: Obtain the augmented matrix

Consider the system of equations (3).

\(\begin{aligned}{c}4.5{x_1} + 3.1{x_2} = 19.249\\1.6{x_1} + 1.1{x_2} = 6.843\end{aligned}\)

\(A{\bf{x}} = b\)can be represented as shown below:

\(\left( {\begin{aligned}{*{20}{c}}{4.5}&{3.1}\\{1.6}&{1.1}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{19.249}\\{6.843}\end{aligned}} \right)\)

Write the augmented matrix\(\left( {\begin{aligned}{*{20}{c}}A&{\bf{b}}\end{aligned}} \right)\)as shown below:

\(\left( {\begin{aligned}{*{20}{c}}{4.5}&{3.1}&{19.249}\\{1.6}&{1.1}&{6.843}\end{aligned}} \right)\)

Convert the augmented matrix into the row-reduced echelon form

Consider the augmented matrix\(\left( {\begin{aligned}{*{20}{c}}{4.5}&{3.1}&{19.249}\\{1.6}&{1.1}&{6.843}\end{aligned}} \right)\).

Use the code in MATLAB to obtain therow-reduced echelon form.

\(\begin{aligned}{l} > > {\rm{ A }} = {\rm{ }}\left( {4.5{\rm{ 3}}{\rm{.1 19}}{\rm{.249; 1}}{\rm{.6 1}}{\rm{.1 6}}{\rm{.843}}} \right);\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{aligned}\)

\(\left( {\begin{aligned}{*{20}{c}}{4.5}&{3.1}&{19.249}\\{1.6}&{1.1}&{6.843}\end{aligned}} \right) \sim \left( {\begin{aligned}{*{20}{c}}1&0&{3.939999622}\\0&1&{0.4900005490}\end{aligned}} \right)\)

Therefore, the solution is\({\bf{x}} = \left( {\begin{aligned}{*{20}{c}}{3.94}\\{0.49}\end{aligned}} \right)\).

The exact solutions of the system of equations (3) are \({x_1} = 3.94\) and \({x_2} = 0.49\).

Obtain the augmented matrix

Consider the system of equations (4).

\(\begin{aligned}{c}4.5{x_1} + 3.1{x_2} = 19.25\\1.6{x_1} + 1.1{x_2} = 6.84\end{aligned}\)

\(A{\bf{x}} = b\)can be represented as shown below:

\(\left( {\begin{aligned}{*{20}{c}}{4.5}&{3.1}\\{1.6}&{1.1}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{19.25}\\{6.84}\end{aligned}} \right)\)

Write theaugmentedmatrix \(\left( {\begin{aligned}{*{20}{c}}A&{\bf{b}}\end{aligned}} \right)\) as shown below:

\(\left( {\begin{aligned}{*{20}{c}}{4.5}&{3.1}&{19.25}\\{1.6}&{1.1}&{6.84}\end{aligned}} \right)\)

Convert the augmented matrix into the row-reduced echelon form

Consider theaugmentedmatrix\(\left( {\begin{aligned}{*{20}{c}}{4.5}&{3.1}&{19.25}\\{1.6}&{1.1}&{6.84}\end{aligned}} \right)\).

Use the code in MATLAB to obtain the row-reducedechelonform.

\(\begin{aligned}{l} > > {\rm{ A }} = {\rm{ }}\left( {4.5{\rm{ 3}}{\rm{.1 19}}{\rm{.25; 1}}{\rm{.6 1}}{\rm{.1 6}}{\rm{.84}}} \right);\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{aligned}\)

\(\left( {\begin{aligned}{*{20}{c}}{4.5}&{3.1}&{19.249}\\{1.6}&{1.1}&{6.843}\end{aligned}} \right) \sim \left( {\begin{aligned}{*{20}{c}}1&0&{2.899999690}\\0&1&{2.000000450}\end{aligned}} \right)\)

Therefore, the solution is\({\bf{x}} = \left( {\begin{aligned}{*{20}{c}}{2.90}\\{2.00}\end{aligned}} \right)\).

The exact solutions of the system of equations (4) are \({x_1} = 2.90\) and \({x_2} = 2.00\).

(b) Step 5: Calculate the relative errors

The exact value of\({x_1}\)in (3) is 3.94, and in (4), it is 2.90.

The relativeerrorin\({x_1}\)is

\(\begin{aligned}{c}\frac{{3.94 - 2.90}}{{3.94}} = \frac{{1.04}}{{3.94}}\\ \approx 0.2639.\end{aligned}\)

In percentage, it is\(26\% \).

The percentage error in\({x_1} = 2.90\)when using the solution of (4) as an approximation for the solution of (3) is\(26\% \).

The exact value of\({x_1}\)in (3) is 3.94, and in (4), it is 2.90.

So, the relativeerrorin\({x_2}\)is

\(\begin{aligned}{c}\frac{{0.49 - 2.00}}{{0.49}} = \frac{{ - 1.51}}{{0.49}}\\ \approx - 3.0816.\end{aligned}\)

In percentage, it is\(308\% \).

The percentage error in \({x_2} = 2.00\) when using the solution of (4) as an approximation for the solution of (3) is \(308\% \).

(c) Step 6: Use the condition number of the coefficient matrix

Consider the coefficient matrix\(A = \left( {\begin{aligned}{*{20}{c}}{4.5}&{3.1}\\{1.6}&{1.1}\end{aligned}} \right)\).

Obtain the condition number of the coefficient matrix by using the MATLAB command shown below:

\(\begin{aligned}{l} > > {\rm{ A }} = {\rm{ }}\left( {4.5{\rm{ 3}}{\rm{.1 19}}{\rm{.249; 1}}{\rm{.6 1}}{\rm{.1 6}}{\rm{.843}}} \right);\\ > > {\rm{ C}} = {\rm{cond}}\left( {\rm{A}} \right)\end{aligned}\)

It gives the output 3362.999.

Thus, the condition number of the coefficient matrix is 3363.